# Will gravity pull together two bodies from the other side of an empty universe?

Lets say that there are only two bodies in the universe, 65 kg each. Other than that the universe is completely empty, no neutrons, no photons, no dark energy/matter, not even neutrinos (that is to make things less complicated. If the loss of other things leads to something like the universe exploding like a bubble at the speed of light or something, you can change these parameters. I'm mainly concerned about gravity here). Those two bodies are placed apart from each other at the distance of the observable universe. Will they start moving into each other? Will they collide? (Optional question: If so, with how much speed will they collide?)

-
Yes they will attract each other. They will only collide if they have zero (or very nearly zero) orbital angular momentum between them. –  Brandon Enright Sep 1 '14 at 0:56
To find the collision speed, you need to know the sizes of the bodies. If they were points, they'd collide at infinite speed. –  Nate Eldredge Sep 1 '14 at 2:43
Those two bodies are placed apart from each other at the distance of the observable universe. By what? If, as specified, the universe is empty except for these two bodies then: (1) before now, they were farther apart and thus, are already moving towards each other or (2) they were closer together and so, have been moving apart. If, now, they are momentarily motionless, then the only option is (2) in which case, their moving apart has ceased and they will now start to move towards each other. –  Alfred Centauri Sep 1 '14 at 2:58
Using the general formula here you can work out the initial gravitational potential energy (almost zero) and the GPE when the two bodies collide (when their distance equals their size). Assuming all that energy is converted to kinetic energy, you can easily compute the velocities, either classically or relativistically. –  Nate Eldredge Sep 1 '14 at 4:01
@DavidHammen Sometimes it is useful to reason about situations that would never arise in reality. For example, nobody would seriously propose locking an actual cat in a box with a poison capsule that may or may not break; but reasoning about such a cat helps us understand the role of an observer in the collapse of a wave function. Likewise, if reasoning about a two-body universe helps us understand celestial mechanics, then there is no reason to dismiss the process as "unscientific". –  David Wallace Sep 1 '14 at 11:18

I assume a steady-state universe and that the bodies have no velocity relative to each other.

Yes, they will eventually collide. Gravity has an effect over any distance, including the ~46 billion light-year radius that constitutes the spherical observable universe (the actual size of the universe may be much larger). Of course, the force will not be very strong over a 100 billion light-year separation, so the bodies would not collide for a very long time. A rough estimate of the time taken would be on the order of billions of years.

EDIT: As pointed out in the comments, the above time estimation was wrong by a over a factor of $10^{20}$. The amount of time taken would be around $10^{38}$ years (100 undecillion years or 100 sextillion years, depending on whether you subscribe to the short scale or long scale). The equation used to find this number can be found here.

-
Billions? At that distance, I'd reckon on something like 10$^{36}$ years. –  Kyle Kanos Sep 1 '14 at 2:15
So this answer (and possibly the question) ignore the expansion of space. In our universe wouldn't they never collide? Wouldn't they actually get farther and farther away from each other? Edit: oops, I see the "no dark energy" part of the question. –  Brandon Enright Sep 1 '14 at 2:27
@KyleKanos: Yep, using this equation I get $3.43\times 10^{36}$ years. –  Nate Eldredge Sep 1 '14 at 2:37
Thank you all for your answers. First of all I won a bet, so thanks to your answers I won't get to spend a whole day doing whatever the other person told me to do. Unfortunately the other person backed away from the bet, so I won't get my prize either. Anyway, if your calculations are correct @NateEldredge the two bodies will never collide. At 3.43×10^36 years practically all their atoms would have decayed. (Didn't account for their speed. If they travel at a percentage of the speed of light, the decay will be a lot slower. Damn.) –  Dimitrios Denton Sep 1 '14 at 3:46
They may never meet if the metric expansion of space exceeds the relative velocity. –  Matt McNabb Sep 1 '14 at 12:45

Yes, they will collide given the initial conditions.

The speed at collision can be calculated. We can presume them to begin with virtually zero gravitational potential energy. We need an assumption of size when they collide. Let's assume a size of 50cm. That way when they collide, the centers will be 1m apart.

$$U = -\frac{GMm}{d}$$ When they collide, the gravitational energy will be $$U = -\frac{(6.67 \times 10^{-11} \frac{Jm}{kg^2})(65kg)^2}{1m}$$ $$U = -2.8 \times 10^{-7} J$$

That's the kinetic energy shared by both when they collide. Each one has half of that energy since they have equal masses. $$v = \sqrt{\frac{2KE}{m}}$$ $$v = \sqrt{\frac{2.8 \times 10^{-7} J}{65kg}}$$ $$v = 6.6 \times 10^{-5} \frac ms$$

That's the speed each has relative to their common center of mass. The time it takes to collide is a more difficult calculation.

-
Another way to get to this result is to note that the objects will be travelling at escape velocity at the time of impact. –  Taemyr Sep 1 '14 at 7:34

If the bodies are initially at rest, then the orbit will be a degenerate ellipsis of finite semi-major axis and eccentricity 1, i.e., a line segment. The semi-major axis $a$ is half the initial distance. Time to collision is half the period $T$. This can be directly derived from Kepler's Third Law.

$$\frac{T^2}{a^3} = \frac{4\pi^2}{GM}$$ $$T = \sqrt{\frac{4\pi^2a^3}{GM}}$$

If we substitute $a = 46\times 10^9~\mathrm{ly}$ (radius of observable universe), $M = 2\times 65~\mathrm{kg}$ and $G = 6.67\times 10^{-11}~\mathrm{Nm}^2/\mathrm{kg}^2$, we get $T = 6.2\times 10^{44}~\mathrm{s}$. Time to collision is thus $\frac{T}{2} = 3.1\times 10^{44}~\mathrm{s}$, which, according to WolframAlpha, is roughly $7\times 10^{26}$ times the age of the universe.

As noted by others, and developed in BowlOfRed's answer, collision speed may be derived by equating gained potential energy and final kinetic energy.

$$\frac{Gm^2}{d} = 2\times \frac{mv^2}{2}$$ $$v = \sqrt{\frac{Gm}{d}}$$

Here $m = 65~\mathrm{kg}$, and $d$ is the final distance, assumed to be much shorter than the initial distance. For, e.g., $d = 1~\mathrm{m}$, we get $v = 7 \times 10^{-5}~\mathrm{m/s} = 70~\mu\mathrm{m/s}$. The relative speed is of course $2v$.

-

The two bodies will collide at high relatavistic speed, it is conceivable that the actual collision velocity is superluminal compared to each other.

With nothing else to interfere the gravitational attraction will be on-axis. I haven't done the math but 10^36 years sounds high - as the attraction forces increase so will the velocity, and the curves are non-linear. It will take a while to get going though. And that's the cosmological "while".

And we have to ask by whose clock are we measuring the speed and the time? stationary clock in the middle (massless, of course) or by clocks inside each object?

-
Nice point to consider. I guess an answer to both cases (stationary massless clock and clocks inside the objects) would be nice, although I think it is fair to say that in both cases it will be a long ass time. :) –  Dimitrios Denton Sep 1 '14 at 3:54
If you do the math, unless I have made a mistake, you'll find that number is not high. It takes a really long time to get going - the gravitational forces at that distance are extremely small. –  Nate Eldredge Sep 1 '14 at 4:03
This is completely wrong. Ignoring general relativity (which I'm not familiar enough with; I'm not even sure the universe described could exist general relativistically), under Newtonian inverse square attraction, bodies even being at infinite distance represent ony finite gravitational potential energy, relative to the position where they are in direct contact. It is this energy that defines the escape velocity, and by time reversal one sees that in described experiment collision speed will equal the escape velocity. Which is extremely small for normal size bodies of 65 kg (not black holes) –  Marc van Leeuwen Sep 1 '14 at 13:23
Yes, this is entirely wrong. $10^{36}$ years is the correct value, speeds will be non-relativistic. –  Kyle Kanos Sep 1 '14 at 14:38

if only gravity is taken into account, then when they reach each other after an exceedingly long time they could pass through each other as there is no electromagnetism. I may have missed the point a little and also as a consequence the objects would sort of disintegrate. also would the answer vary depending on the size of the universe?

-